Cube Numbers as Sum of Sequence of Cubes

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Theorem

\(\displaystyle 16 \, 830^3\) \(=\) \(\displaystyle \sum_{k \mathop = 1134}^{2133} k^3\)
\(\displaystyle \) \(=\) \(\displaystyle 1134^3 + 1135^3 + \cdots + 2133^3\)


Proof

\(\displaystyle \sum_{k \mathop = 1134}^{2133} k^3\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^{2133} k^3 - \sum_{k \mathop = 1}^{1133} k^3\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2133^2 \paren {2133 + 1}^2} 4 - \frac {1133^2 \paren {1133 + 1}^2} 4\) Sum of Sequence of Cubes
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {3^3 \times 79}^2 \paren {2 \times 11 \times 97}^2} 4 - \frac {\paren {11 \times 103}^2 \paren {2 \times 3^4 \times 7}^2} 4\) extracting Prime Decomposition
\(\displaystyle \) \(=\) \(\displaystyle 3^6 \times 11^2 \paren {\paren {79 \times 97}^2 - \paren {103 \times 3 \times 7}^2}\) factorising
\(\displaystyle \) \(=\) \(\displaystyle 3^6 \times 11^2 \paren {7663^2 - 2163^2}\) factorising
\(\displaystyle \) \(=\) \(\displaystyle 3^6 \times 11^2 \paren {7663 + 2163} \times \paren {7663 - 2163}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle 3^6 \times 11^2 \times 9826 \times 5500\)
\(\displaystyle \) \(=\) \(\displaystyle 3^6 \times 11^2 \times \paren {2 \times 17^3} \times \paren {2^2 \times 5^3 \times 11}\)
\(\displaystyle \) \(=\) \(\displaystyle 2^3 \times 3^6 \times 5^3 \times 11^3 \times 17^3\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {2 \times 3^2 \times 5 \times 11 \times 17}^3\)
\(\displaystyle \) \(=\) \(\displaystyle 16 \, 830^3\)

$\blacksquare$


Sources